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Within a rectangular courtyard of length 60 feet, a graveled path,

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Intern
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Within a rectangular courtyard of length 60 feet, a graveled path,  [#permalink]

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New post 24 Sep 2018, 09:25
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Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :

A 24
B 30
C 40
D 45
E 54
Intern
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Within a rectangular courtyard of length 60 feet, a graveled path,  [#permalink]

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New post 24 Sep 2018, 10:12
Being x the width of the rectangle, we have the following equation:

\(2[2(60*3)+2*3(x-6)] + 984 = 2[2(60*6)+2*6(x-12)]\)

where \(2(60*3)\) is the area of the path next to the longer side (60 ft long and 3 ft wide), and \(2*3(x-6)\) the area of the path next to the shorter side, avoiding to count the area from the corner twice. Then we need to repeat the same idea with the bigger path. Both areas are multiplied by 2 because we are measuring the cost, which is Rs 2 per sq ft. We have to consider that the path with double size is Rs 984 more expensive, so we add this into our equation. Then, we can solve the equation:

\(2[360+6x-36]+984=2(720+12x-144)\)
\(360+6x-36+492=720+12x-144\)
\(324+6x+492=576+12x\)
\(816-576=6x\)
\(240=6x\)
\(x=40\)

Thus, C is the correct answer.
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Re: Within a rectangular courtyard of length 60 feet, a graveled path,  [#permalink]

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New post 24 Sep 2018, 10:33
sultanatehere wrote:
Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :

A 24
B 30
C 40
D 45
E 54


A = Area of outer rectangle = 60 x Width
Ax = Area of inner rectangle with width of 3 = 54 x (Width-6)
Ay = Area of inner rectangle with twice the width, i.e 6 = 48 x (Width - 12)

Area of gravel with normal width = A-Ax
Area of gravel with twice width = A-Ay

Given, cost of gravelling twice width - cost of gravelling normal width = 984.
Given, cost of gravelling = Rs2/sq ft.

2(A-Ay) - 2(A-Ax) = 984.
A-Ay-A+Ax=492.
Ax-Ay=492.

On substituting Ax and Ay from above and solving the above equation, we get width=40.

Hence C.

Cheers!
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Re: Within a rectangular courtyard of length 60 feet, a graveled path,  [#permalink]

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New post 24 Sep 2018, 12:48
Tried to solve using an Ans choices and got lucky with the 1st option :cool:
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Re: Within a rectangular courtyard of length 60 feet, a graveled path,  [#permalink]

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New post 27 Sep 2018, 16:20
sultanatehere wrote:
Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :

A 24
B 30
C 40
D 45
E 54


We can let w = the width of the courtyard. Therefore, the courtyard, including the graveled path, has an area of 60w sq ft. Excluding the path, the area of the courtyard would be (60 - 2(3))(w - 2(3)) = 54(w - 6) = 54w - 324 sq ft. So the path alone has an area of 60w - (54w - 324) = 6w + 324 sq ft. Since the cost of graveling the path is Rs 2 per sq ft, the cost of graveling the path is Rs 12w + 648.

If the path had been twice as wide, then the courtyard, including the path, would still have an area of 60w sq ft. However, excluding the path, the area of the courtyard would be (60 - 2(6))(w - 2(6)) = 48(w - 12) = 48w - 576 sq ft. So the path alone would have an area of 60w - (48w - 576) = 12w + 576 sq ft, and the cost of the path would be Rs 24w + 1152. We are told that this would cost Rs 984 more, so we can set up the following equation to solve for w:

12w + 648 + 984 = 24w + 1152

12w + 1632 = 24w + 1152

480 = 12w

40 = w

Answer: C
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Re: Within a rectangular courtyard of length 60 feet, a graveled path, &nbs [#permalink] 27 Sep 2018, 16:20
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